在实数范围内分解因式:(1)x²+x-1 (2)x²-2x-6 (3)2x²+8x-7 (

1个回答

  • (1)x²+x-1=x²+x+1/4-5/4=(x+1/2)²-5/4=(x+1/2+√5/2)(x+1/2-√5/2)

    (2)x²-2x-6=x²-2x+1-7=(x-1)²-7=(x-1+√7)(x-1-√7)

    (3)2x²+8x-7=2(x²+4x-7/2)=2(x²+4x+4-15/2)=2[(x+2)²-15/2]=2[(x+2)²-15/2]=2(x+2+√30/2)(x+2-√30/2)

    (4)x²-5xy+3y²=x²-5xy+25/4y²-13/4y²=(x-5/2y)²-13/4y²=(x-5/2y+√13/2y)(x-5/2y-√13/2y)

    (5)-2x²-3x+6=-2(x²+3/2x-3)=-2(x²+3/2x+9/16-57/16)=-2[(x+3/4)²-57/16]=-2(x+3/4+√57/4)(x+3/4-√57/4)

    (6)4x²y²+xy-1=4x²y²+xy+1/16-17/16=(2xy+1/4)²-17/16=(2xy+1/4+√17/4)(2xy+1/4-√17/4)